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A holding session with B and calling on session with C. A transfer B to C by implementing code below. How to get info of B in C, i tried listening ua.on('newMessage', cb) and adding ua.sendMessage(currentSession?.remote_identity.uri, 'dataOfB') in A but ua.on('newMessage', cb) only fired in A. How to C detect call transfered successfully or get message from A. Thanks all!

    replaces: currentSession,
    mediaConstraints: { audio: true, video: false },
     pcConfig: {
     rtcpMuxPolicy: 'negotiate',
    },
  };
  holdSession?.refer(currentSession?.remote_identity.uri, options);
  currentSession?.terminate();
  holdSession?.terminate();
Yaokaoya
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1 Answers1

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As stated in their docs:

A REFER method implicitly generates a subscription to the state of the refer. The NOTIFY requests received from the peer are processed and presented in form of events by JsSIP.RTCSession.ReferSubscriber.

The documentation on these events can be found here.

Take a look at this other answer for a working example.

juxnpxblo
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    Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – Community Nov 17 '22 at 08:21